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Case Study Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

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Here we are providing case study questions for class 11 maths. In this article, we are sharing Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations case study questions. All case study questions of class 11 maths are solved so that students can check their solutions after attempting questions.

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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CBSE Class 11 Maths – Chapter 5 Complex Numbers and Quadratic Equations- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sets : Notes and Study Materials -pdf

• Concepts of  Complex Numbers and Quadratic Equations
• Complex Numbers and Quadratic Equations Master File
• Complex Numbers and Quadratic Equations Revision Notes
• R D Sharma Solution of Complex Numbers
• R D Sharma Solution of Quadratic Equations
• NCERT Solution  Complex Numbers and Quadratic Equations
• NCERT  Exemplar Solution Complex Numbers and Quadratic Equations
• Complex Numbers and Quadratic Equations : Solved Example 1

CBSE Class 11 Maths Notes Chapter 5 Complex Numbers and Quadratic Equations

Imaginary Numbers The square root of a negative real number is called an imaginary number, e.g. √-2, √-5 etc. The quantity √-1 is an imaginary unit and it is denoted by ‘i’ called Iota.

Integral Power of IOTA (i) i = √-1, i 2  = -1, i 3  = -i, i 4  = 1 So, i 4n+1  = i, i 4n+2  = -1, i 4n+3  = -i, i 4n  = 1

• For any two real numbers a and b, the result √a × √b : √ab is true only, when atleast one of the given numbers i.e. either zero or positive. √-a × √-b ≠ √ab So, i 2  = √-1 × √-1 ≠ 1
• ‘i’ is neither positive, zero nor negative.
• i n  + i n+1  + i n+2  + i n+3  = 0

Complex Number A number of the form x + iy, where x and y are real numbers, is called a complex number, x is called real part and y is called imaginary part of the complex number i.e. Re(Z) = x and Im(Z) = y.

Purely Real and Purely Imaginary Complex Number A complex number Z = x + iy is a purely real if its imaginary part is 0, i.e. Im(z) = 0 and purely imaginary if its real part is 0 i.e. Re (z) = 0.

Equality of Complex Number Two complex numbers z 1  = x 1  + iy 1  and z 2  = x 2  + iy 2  are equal, iff x 1  = x 2  and y 1  = y 2  i.e. Re(z 1 ) = Re(z 2 ) and Im(z 1 ) = Im(z 2 ) Note: Order relation “greater than’’ and “less than” are not defined for complex number.

Algebra of Complex Numbers Addition of complex numbers Let z 1  = x 1  + iy 1  and z 2  = x 2  + iy 2  be any two complex numbers, then their sum defined as z 1  + z 2  = (x 1  + iy 1 ) + (x 2  + iy 2 ) = (x 1  + x 2 ) + i (y 1  + y 2 )

Properties of Addition

• Commutative: z 1  + z 2  = z 2  + z 1
• Associative: z 1  + (z 2  + z 3 ) = (z 1  + z 2 ) + z 3
• Additive identity z + 0 = z = 0 + z Here, 0 is additive identity.

Subtraction of complex numbers Let z 1  = (x 1  + iy 1 ) and z 2  = (x 2  + iy 2 ) be any two complex numbers, then their difference is defined as z 1  – z 2  = (x 1  + iy 1 ) – (x 2  + iy 2 ) = (x 1  – x 2 ) + i(y 1  – y 2 )

Multiplication of complex numbers Let z 1  = (x 1  + iy 1 ) and z 2  = (x 2  + iy 2 ) be any two complex numbers, then their multiplication is defined as z 1 z 2  = (x 1  + iy 1 ) (x 2  + iy 2 ) = (x 1 x 2  – y 1 y 2 ) + i (x 1 y 2  + x 2 y 1 )

Properties of Multiplication

• Commutative: z 1 z 2  = z 2 z 1
• Associative: z 1 (z 2 z 3 ) = (z 1 z 2 )z 3
• Multiplicative identity: z . 1 = z = 1 . z Here, 1 is multiplicative identity of an element z.
• Multiplicative inverse: For every non-zero complex number z, there exists a complex number z 1  such that z . z 1  = 1 = z 1  . z
• Distributive law: z 1 (z 2  + z 3 ) = z 1 z 2  + z 1 z 3

Conjugate of Complex Number Let z = x + iy, if ‘i’ is replaced by (-i), then said to be conjugate of the complex number z and it is denoted by  z ¯ , i.e.  z ¯  = x – iy

Modulus of a Complex Number Let z = x + iy be a complex number. Then, the positive square root of the sum of square of real part and square of imaginary part is called modulus (absolute values) of z and it is denoted by |z| i.e. |z| =  x 2 + y 2 −−−−−−√ It represents a distance of z from origin in the set of complex number c, the order relation is not defined i.e. z 1  > z 2  or z 1  < z 2  has no meaning but |z 1 | > |z 2 | or |z 1 |<|z 2 | has got its meaning, since |z 1 | and |z 2 | are real numbers.

Argand Plane Any complex number z = x + iy can be represented geometrically by a point (x, y) in a plane, called argand plane or gaussian plane. A purely number x, i.e. (x + 0i) is represented by the point (x, 0) on X-axis. Therefore, X-axis is called real axis. A purely imaginary number iy i.e. (0 + iy) is represented by the point (0, y) on the y-axis. Therefore, the y-axis is called the imaginary axis.

Argument of z is not unique, general value of the argument of z is 2nπ + θ, but arg(0) is not defined. The unique value of θ such that -π < θ ≤ π is called the principal value of the amplitude or principal argument.

Principal Value of Argument

• if x > 0 and y > 0, then arg(z) = θ
• if x < 0 and y > 0, then arg(z) = π – θ
• if x < 0 and y < 0, then arg(z) = -(π – θ)
• if x > 0 and y < 0, then arg(z) = -θ

Polar Form of a Complex Number If z = x + iy is a complex number, then z can be written as z = |z| (cosθ + isinθ), where θ = arg(z). This is called polar form. If the general value of the argument is θ, then the polar form of z is z = |z| [cos (2nπ + θ) + isin(2nπ + θ)], where n is an integer.

Square Root of a Complex Number

Solution of a Quadratic Equation The equation ax 2  + bx + c = 0, where a, b and c are numbers (real or complex, a ≠ 0) is called the general quadratic equation in variable x. The values of the variable satisfying the given equation are called roots of the equation.

The quadratic equation ax 2  + bx + c = 0 with real coefficients has two roots given by  − b + √ D 2 a  and  − b − √ D 2 a , where D = b 2  – 4ac, called the discriminant of the equation.

Note: (i) When D = 0, roots ore real and equal. When D > 0 roots are real and unequal. Further If a,b, c ∈ Q and D is perfect square, then the roots of quadratic equation are real and unequal and if a, b, c ∈ Q and D is not perfect square, then the roots are irrational and occur in pair. When D < 0, roots of the equation are non real (or complex).

(ii) Let α, β be the roots of quadratic equation ax 2  + bx + c = 0, then sum of roots α + β =  − b a  and the product of roots αβ =  c a .

CBSE Class 11 Maths Chapter-5 Important Questions

1 Marks Questions

1. Evaluate i -39

4. Evaluate (1+ i) 4

6. Express in the form of a + ib. (1+3i) -1

9. Find the conjugate of – 3i – 5.

Ans.  Let z = 3i – 5

Ans.  z 1  z 2  = (2 – i)(-2 + i)

11. Express in the form of a + ib (3i-7) + (7-4i) – (6+3i) + i 23

13. Solve for x and y, 3x + (2x-y) i= 6 – 3i

Ans.  3x = 6

2x – y = – 3

2 × 2 – y = – 3

– y = – 3 – 4

14. Find the value of 1+i 2  + i 4  + i 6  + i 8  + —- + i 20

15. Multiply 3-2i by its conjugate.

Ans.Let z = 3 – 2i

16. Find the multiplicative inverse 4 – 3i.

Ans.  Let z = 4 – 3i

17. Express in term of a + ib

19. If 1, w, w 2  are three cube root of unity, show that (1 – w + w 2 ) (1 + w – w 2 ) = 4

Ans .(1 – w + w 2 ) (1 + w – w 2 )

(1 + w 2  – w) (1 + w – w 2 )

21. Write the real and imaginary part 1 – 2i 2

Ans.  Let z = 1 – 2i 2

=1 – 2 (-1)

Re (z) = 3, Im (z) = 0

22. If two complex number z 1 , z 2  are such that |z 1 | = |z 2 |, is it then necessary that z 1  = z 2

Ans. Let z 1  = a + ib

24. Find the number of non zero integral solution of the equation |1-i| x  = 2 x

Which is false no value of x satisfies.

25. If (a + ib) (c + id) (e + if) (g + ih) = A + iB then show that

4 Marks Questions

taking conjugate both side

x 2  + y 2  = 1

For purely real

Taking conjugate both side

a 2  + b 2  = 1

Ans. z 1  + z 2  + 1 = 2 – i + 1+ i + 1 = 4

8.If (p + iq) 2  = x + iy Prove that (p 2  + q 2 ) 2  = x 2  + y 2

Ans.( p + iq) 2  = x + iy (i)

(p – iq) 2  = x –iy (ii)

Ans .i 25  + (1+3i) 3

16.For what real value of x and y are numbers equal (1+i) y 2  + (6+i) and (2+i) x

Ans. (1+i) y 2  + (6 + i) = (2 + i) x

y 2  + iy 2  + 6 + i = 2x + xi

(y 2  + 6) + (y 2  + 1) i = 2x + xi

y 2  + 6 = 2x

y 2  + 1 = x

y  2  = x – 1

x – 1 + 6 = 2x

19.Find the real values of x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i

(x – iy) (3 + 5i) = – 6 + 24i

3x + 5xi – 3yi – 5yi 2  = – 6 + 24i

6 Marks Questions

Since  Re (z) < o, and Im (z) > o

3.Find two numbers such that their sum is 6 and the product is 14.

Ans. Let x and y be the no.

Complex Numbers and Quadratic Equations Class 11 MCQs Questions with Answers

Question 1. The value of √(-16) is (a) -4i (b) 4i (c) -2i (d) 2i

Answer: (b) 4i Hint: Given, √(-16) = √(16) × √(-1) = 4i {since i = √(-1) }

Question 2. The value of √(-144) is (a) 12i (b) -12i (c) ±12i (d) None of these

Answer: (a) 12i Hint: Given, √(-144) = √{(-1) × 144} = √(-1) × √(144) = i × 12 {Since √(-1) = i} = 12i So, √(-144) = 12i

Question 3: The value of √(-25) + 3√(-4) + 2√(-9) is (a) 13i (b) -13i (c) 17i (d) -17i

Answer: (c) 17i Hint: Given, √(-25) + 3√(-4) + 2√(-9) = √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9} = √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9} = 5i + 3×2i + 2×3i {since √(-1) = i} = 5i + 6i + 6i = 17i So, √(-25) + 3√(-4) + 2√(-9) = 17i

Question 4. if z lies on |z| = 1, then 2/z lies on (a) a circle (b) an ellipse (c) a straight line (d) a parabola

Answer: (a) a circle Hint: Let w = 2/z Now, |w| = |2/z| => |w| = 2/|z| => |w| = 2 This shows that w lies on a circle with center at the origin and radius 2 units.

Question 5. If ω is an imaginary cube root of unity, then (1 + ω – ω²) 7 equals (a) 128 ω (b) -128 ω (c) 128 ω² (d) -128 ω²

Answer: (d) -128 ω² Hint: Given ω is an imaginary cube root of unity. So 1 + ω + ω² = 0 and ω³ = 1 Now, (1 + ω – ω²) 7 = (-ω² – ω²) 7 ⇒ (1 + ω – ω 2 ) 7 = (-2ω 2 ) 7 ⇒ (1 + ω – ω 2 ) 7 = -128 ω 14 ⇒ (1 + ω – ω 2 ) 7 = -128 ω 12 × ω 2 ⇒ (1 + ω – ω 2 ) 7 = -128 (ω 3 ) 4 ω 2 ⇒ (1 + ω – ω 2 ) 7 = -128 ω 2

Question 6. The least value of n for which {(1 + i)/(1 – i)} n is real, is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (b) 2 Hint: Given, {(1 + i)/(1 – i)} n = [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}] n = [{(1 + i)²}/{(1 – i²)}] n = [(1 + i² + 2i)/{1 – (-1)}] n = [(1 – 1 + 2i)/{1 + 1}] n = [2i/2] n = i n Now, in is real when n = 2 {since i2 = -1 } So, the least value of n is 2

Question 7. Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z = (a) -2√3 + 2i (b) 2√3 + 2i (c) 2√3 – 2i (d) -√3 + i

Answer: (a) -2√3 + 2i Hint: Let z = r(cos θ + i × sin θ) Then r = 4 and θ = 5π/6 So, z = 4(cos 5π/6 + i × sin 5π/6) ⇒ z = 4(-√3/2 + i/2) ⇒ z = -2√3 + 2i

Question 8: The value of i -999 is (a) 1 (b) -1 (c) i (d) -i

Answer: (c) i Hint: Given, i -999 = 1/i 999 = 1/(i 996 × i³) = 1/{(i 4 ) 249 × i 3 } = 1/{1 249 × i 3 } {since i 4 = 1} = 1/i 3 = i 4 /i 3 {since i 4 = 1} = i So, i -999 = i

Question 9. Let z 1 and z 2 be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z 1 and z 1 form an equilateral triangle. Then (a) a² = b (b) a² = 2b (c) a² = 3b (d) a² = 4b

Answer: (c) a² = 3b Hint: Given, z 1 and z 2 be two roots of the equation z² + az + b = 0 Now, z 1 + z 2 = -a and z 1 × z 2 = b Since z 1 and z 2 and z 3 from an equilateral triangle. ⇒ z 1 2 + z 2 2 + z 3 2 = z 1 × z 2 + z 2 × z 3 + z 1 × z 3 ⇒ z 1 2 + z 2 2 = z 1 × z 2 {since z 3 = 0} ⇒ (z 1 + z 2 )² – 2z 1 × z 2 = z 1 × z 2 ⇒ (z 1 + z 2 )² = 2z 1 × z 2 + z 1 × z 2 ⇒ (z 1 + z 2 )² = 3z 1 × z 2 ⇒ (-a)² = 3b ⇒ a² = 3b

Question 10: The complex numbers sin x + i cos 2x are conjugate to each other for (a) x = nπ (b) x = 0 (c) x = (n + 1/2) π (d) no value of x

Answer: (d) no value of x Hint: Given complex number = sin x + i cos 2x Conjugate of this number = sin x – i cos 2x Now, sin x + i cos 2x = sin x – i cos 2x ⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part} ⇒ tan x = 1 and tan 2x = 1 Now both of them are not possible for the same value of x. So, there exist no value of x

Question 11. The curve represented by Im(z²) = k, where k is a non-zero real number, is (a) a pair of striaght line (b) an ellipse (c) a parabola (d) a hyperbola

Answer: (d) a hyperbola Hint: Let z = x + iy Now, z² = (x + iy)² ⇒ z² = x² – y² + 2xy Given, Im(z²) = k ⇒ 2xy = k ⇒ xy = k/2 which is a hyperbola.

Question 12. The value of x and y if (3y – 2) + i(7 – 2x) = 0 (a) x = 7/2, y = 2/3 (b) x = 2/7, y = 2/3 (c) x = 7/2, y = 3/2 (d) x = 2/7, y = 3/2

Answer: (a) x = 7/2, y = 2/3 Hint: Given, (3y – 2) + i(7 – 2x) = 0 Compare real and imaginary part, we get 3y – 2 = 0 ⇒ y = 2/3 and 7 – 2x = 0 ⇒ x = 7/2 So, the value of x = 7/2 and y = 2/3

Question 13. Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary (a) θ = nπ ± π/2 where n is an integer (b) θ = nπ ± π/3 where n is an integer (c) θ = nπ ± π/4 where n is an integer (d) None of these

Answer: (b) θ = nπ ± π/3 where n is an integer Hint: Given, (3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ) (3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. 1 Now, equation 1 is imaginary if 3 – 4sin² θ = 0 ⇒ 4sin² θ = 3 ⇒ sin² θ = 3/4 ⇒ sin θ = ±√3/2 ⇒ θ = nπ ± π/3 where n is an integer

Question 14. If {(1 + i)/(1 – i)} n = 1 then the least value of n is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (d) 4 Hint: Given, {(1 + i)/(1 – i)} n = 1 ⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}] n = 1 ⇒ [{(1 + i)²}/{(1 – i²)}] n = 1 ⇒ [(1 + i² + 2i)/{1 – (-1)}] n = 1 ⇒ [(1 – 1 + 2i)/{1 + 1}] n = 1 ⇒ [2i/2] n = 1 ⇒ i n = 1 Now, i n is 1 when n = 4 So, the least value of n is 4

Question 15. If arg (z) < 0, then arg (-z) – arg (z) = (a) π (b) -π (c) -π/2 (d) π/2

Answer: (a) π Hint: Given, arg (z) < 0 Now, arg (-z) – arg (z) = arg(-z/z) ⇒ arg (-z) – arg (z) = arg(-1) ⇒ arg (-z) – arg (z) = π {since sin π + i cos π = -1, So arg(-1) = π}

Question 16. if x + 1/x = 1 find the value of x 2000 + 1/x 2000 is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (c) -1 Hint: Given x + 1/x = 1 ⇒ (x² + 1) = x ⇒ x² – x + 1 = 0 ⇒ x = {-(-1) ± √(1² – 4 × 1 × 1)}/(2 × 1) ⇒ x = {1 ± √(1 – 4)}/2 ⇒ x = {1 ± √(-3)}/2 ⇒ x = {1 ± √(-1)×√3}/2 ⇒ x = {1 ± i√3}/2 {since i = √(-1)} ⇒ x = -w, -w² Now, put x = -w, we get x 2000 + 1/x 2000 = (-w) 2000 + 1/(-w) 2000 = w 2000 + 1/w 2000 = w 2000 + 1/w 2000 = {(w³) 666 × w²} + 1/{(w³) 666 × w²} = w² + 1/w² {since w³ = 1} = w² + w³ /w² = w² + w = -1 {since 1 + w + w² = 0} So, x 2000 + 1/x 2000 = -1

Question 17. The value of √(-144) is (a) 12i (b) -12i (c) ±12i (d) None of these

Answer: (a) 12i Hint: Given, √(-144) = √{(-1)×144} = √(-1) × √(144) = i × 12 {Since √(-1) = i} = 12i So, √(-144) = 12i

Question 18. If the cube roots of unity are 1, ω, ω², then the roots of the equation (x – 1)³ + 8 = 0 are (a) -1, -1 + 2ω, – 1 – 2ω² (b) – 1, -1, – 1 (c) – 1, 1 – 2ω, 1 – 2ω² (d) – 1, 1 + 2ω, 1 + 2ω²

Answer: (c) – 1, 1 – 2ω, 1 – 2ω² Hint: Note that since 1, ω, and ω² are the cube roots of unity (the three cube roots of 1), they are the three solutions to x³ = 1 (note: ω and ω² are the two complex solutions to this) If we let u = x – 1, then the equation becomes u³ + 8 = (u + 2)(u² – 2u + 4) = 0. So, the solutions occur when u = -2 (giving -2 = x – 1 ⇒ x = -1), or when: u² – 2u + 4 = 0, which has roots, by the Quadratic Formula, to be u = 1 ± i√3 So, x – 1 = 1 ± i√3 ⇒ x = 2 ± i√3 Now, x³ = 1 when x³ – 1 = (x – 1)(x² + x + 1) = 0, giving x = 1 and x² + x + 1 = 0 ⇒ x = (-1 ± i√3)/2 If we let ω = (-1 – i√3)/2 and ω₂ = (-1 + i√3)/2 then 1 – 2ω and 1 – 2ω² yield the two complex solutions to (x – 1)³ + 8 = 0 So, the roots of (x – 1)³ + 8 are -1, 1 – 2ω, and 1 – 2ω²

Question 19. (1 – w + w²)×(1 – w² + w 4 )×(1 – w 4 + w 8 ) × …………… to 2n factors is equal to (a) 2 n (b) 2 2n (c) 2 3n (d) 2 4n

Answer: (b) 2 2n Hint: Given, (1 – w + w²)×(1 – w² + w 4 )×(1 – w 4 + w 8 ) × …………… to 2n factors = (1 – w + w 2 )×(1 – w 2 + w )×(1 – w + w 2 ) × …………… to 2n factors {Since w 4 = w, w 8 = w 2 } = (-2w) × (-2w²) × (-2w) × (-2w²)× …………… to 2n factors = (2² w³)×(2² w³)×(2² w³) …………… to 2n factors = (2²) n {since w³ = 1} = 2 2n

Question 20. The modulus of 5 + 4i is (a) 41 (b) -41 (c) √41 (d) -√41

Answer: (c) √41 Hint: Let Z = 5 + 4i Now modulus of Z is calculated as |Z| = √(5² + 4²) ⇒ |Z| = √(25 + 16) ⇒ |Z| = √41 So, the modulus of 5 + 4i is √41

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Important Questions Class 11 Maths Chapter 5

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Important Questions Class 11 Mathematics Chapter 5

Important questions for cbse class 11 mathematics chapter 5 – complex numbers and quadratic equations.

Important Questions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations are easily accessible here on the Extramarks website. These Chapter 5 Class 11 Mathematics Important Questions are extremely beneficial to students as they come with accurate solutions.

Subject matter experts prepare these answers in step-by-step solutions so that students can easily grasp the complex concepts of Mathematics.

To improve their exam scores, students should review these Important Questions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations with solutions.

CBSE Class 11 Mathematics Chapter-5 Important Questions

Study important questions for class 11 mathematics chapter 5 – complex numbers and quadratic equations.

Following are some examples of I mportant Questions in Class 11 Mathematics Chapter 5.

Click the below link to access the Chapter 5 Class 11 Mathematics Important Questions.

Very Short Answer Questions: (1 Mark)

Q.1 Evaluate the value of i −39 .

Ans: Let us solve the given expression further –

i −39 =i− 38−1

⇒i −39 =(i 2 ) −19 . 1 i

⇒i −39 =(−1) −19 . 1 i

⇒i −39 =−1 1 i x i i

Q.2 Evaluate the expression (1+i) 4

Ans: Let us simplify the given expression –

(1+i) 4 =[(1+i) 2 ] 2

⇒(1+i2+i 2 ) 2

⇒(1+i2−1) 2

Hence, (1+i) 4 =−4+0i

Q.3 Find the conjugate of −3i−5.

Ans: Let us consider a complex number 

Therefore, the conjugate will be 

Q.4 Express in the form of a+ib:(3i−7)+(7−4i)−(6+3i)+i 23 .

(3i−7)+(7−4i)−(6+3i)+i 23 =3i−7+7−4i−6−3i+i 23

⇒−4i−6+i 22+1

⇒−4i−6+(i 2 ) 11 .i

⇒−4i−6+(−1) 11 .i

Therefore, (3i−7)+(7−4i)−(6+3i)+i 23 =−6−5i.

Q.5 Solve for x and y, 3x+(2x−y)i=6−3i.

Ans: We will equate the real part of the right-hand side with the real part of the left-hand side. Similarly, we will equate their imaginary parts as well.

3x=6 and 2x−y=−3

⇒x=2 and ⇒2(2)−y=−3

Hence, x=2 and y=7

Q.6 Find the value of 1+i 2 +i 4 +i 6 +i 8 +……+i 20 .

1+i 2 +i 4 +i 6 +i 8 +……+i 20 =1−1+(i 2 ) 2 +(i 2 ) 3 +(i 2 ) 4 +……+(i 2 ) 10

=(−1) 2 +(−1) 3 +(−1) 4 +……+(−1) 10

=1−1+1−1+……+1

Therefore, 1+i 2 +i 4 +i 6 +i 8 +……+i 20 =1

Q.7 Multiply 3−2i by its conjugate.

Ans: Let there be a complex number z=3−2i

Hence, its conjugate will be z¯=3+2i

Therefore, the product of the complex number with its conjugate will be –

zz¯=(3−2i)(3+2i)

⇒9+6i−6i−4i2

Hence, (3−2i)(3+2i)=13

Q.8 Write the real and imaginary parts 1−2i 2 .

1−2i 2 =1−2(−1)

Hence, the real and imaginary parts of 1−2i 2 are 3 and 0.

Long Answer Questions: (4 Marks)

Q.1 For what real value of x and y are numbers equal (1+i)y 2 +(6+i) and (2+i)x

Ans: Let us equate both the numbers as –

(1+i)y 2 +(6+i)=(2+i)x

⇒(y 2 +6)+i(y 2 +1)=2x+ix

After solving both the equations obtained, we obtain–

x=5 and y=±2

Q.2 Find the real numbers x and y if (x−iy)(3+5i) is the conjugate of −6−24i.

Ans: We know that the conjugate of the given complex number z=−6−24i will be z¯=−6+24i.

Now, let us simplify the expression (x−iy)(3+5i)-

Hence, (x−iy)(3+5i)=3x+5xi−3yi+5y

⇒(3x+5y)+i(5x−3y)

Now, we will compare the values of the expression with the conjugate of the complex number.

−6+24i=(3x+5y)+i(5x−3y)

Thus, we have –

−6=3x+5y and 24=5x−3y.

Now, we will solve both equations for x and y.

Hence, we have 51=−17y

After substituting the value in the first equation we get –

Therefore, the real numbers x and y are 3 and −3.

Long Answer Questions: (6 Marks)

Q.1 If z=x+iy and w= 1-iz z-i . Show that | w|=1⇒z is purely real.

Ans: Given we have z=x+iy

Hence, we have

w= 1-i (x+iy) x+iy-i

w = 1+y−ix x+i(y−1)

1+y−ix x+i(y−1) = 1

|1+y−ix| |x+i(y−1)| =1

(1+y ) 2 + x 2 x 2 + (y-1 ) 2 =1 

⇒(1+y) 2 +x 2 =x 2 +(y−1) 2

⇒1+y 2 +2y=y 2 +1−2y

Hence, z=x+0i implying that z is purely real.

Q.2  Find two numbers such that their sum is 6

 and the product is 14

Let us consider x and y as the two numbers.

Hence, we have –

⇒6x−x 2 =14

⇒x 2 −6x+14=0

x= 6 ± 36-56 2

x= 6 ± 20i 2

Hence y= 6- ( 3 ± 5i )

Class 11 Mathematics Chapter 5 Important Question

Extramarks provide Chapter 5 Class 11 Mathematics Important Questions . Complex numbers and Quadratic Equations in Class 11 explains how to solve sums using complex numbers.

Topics Covered in the Chapter Complex Numbers and Quadratic Equations

Before going through Mathematics Class 11 Chapter 5 Important Questions , students can review the topics covered in this chapter given below.

• The real number and an imaginary number
• Complex number definition
• Integral powers of iota
• Purely real and purely imaginary complex numbers.
• Complex number equality
• Algebra in complex numbers – addition, subtraction, multiplication, and division
• Conjugate of a complex number.
• Modulus in a complex number
• Argand plane

Before understanding the concept of the complex number, one must first understand the meaning of the real number and the imaginary number.

Important Definitions

Real Number : A number on the number line that is in the form of a positive, negative, rational, irrational, zero, fraction, integer, etc., is called a real number. 

For example: 34, -3, 7 , 22/67, 0 

Imaginary Number : The numbers except the real numbers is called imaginary number. It is the root of a negative number.

For example: -67, -23/92 etc

Complex Number : A complex number is defined as a number that can be expressed in the form of a + ib.

Here, a and b are real numbers and i is iota which will be discussed.

The value of iota is R-1. 

Therefore, z (complex number) = a + ib where a is the real part, and ib is the imaginary part. 

The Integral Powers of Lota

i³ = i².i = (-1). i = -i

i⁴ = i².i² = (-1) (-1) = 1

Now we can generalise it, therefore;

i 4n+1   = i

i 4n+2 = = -1

i 4n+3 = -i

It can also be generalised in this form,

If n= even integer then, i n = (-1) n/2

And, if n= odd integer then, i n   = (-1) (n−1)/2

What is a Purely Real Complex Number and a Purely Imaginary Complex Number?

When the imaginary part of a complex number is zero, it is referred to as a purely real complex number.

When the real part of a complex number is zero, it is referred to as a purely imaginary complex number.

Complex Number Equality

Two complex numbers are taken, z1 and z2

z 1 = a 1 + ib 1

z 2 = a 2 + ib 2

If z 1 = z 2

i.e. a 1 + ib 1 = a 2 + ib 2

then, a 1 = a 2 and b 1 = b 2

Therefore, Re( z1 ) = Re( z2 )

And Im( z1 ) = Im (z2 )

Algebra in Complex Number

Let, z1 = a1 + ib1 and, z2 = a2+ ib2

Adding both the complex numbers we get,

z 1 + z 2 = (a 1 + ib 1 ) + (a 2 + ib 2 )

=  (a 1 + a 2 ) + i(b 1 + b 2 )  

Properties:

• Commutative: z 1 + z 2 = z 2 + z 1
• Associative: (z 1 + z) + z 3 = z 1 + (z 2 + z 3 )
• Additive identity: 0+z = z+0 = z

Solved Examples of Complex Numbers

Q1. Express (5 – 3i)³ in the form a + ib.

Ans:  We have, (5 – 3i)³ 

= 5³ – 3 × 5²× (3i) + 3 × 5 (3i)² – (3i)³ 

= 125 – 225i – 135 + 27i 

= – 10 – 198i

Q2. Simplify

• a) 16i + 10i(3-i)
• b) (7i)(5i)
• c) 11i + 13i – 2i

= 16i + 10i(3) + 10i (-i)

= 16i +30i – 10 i2

= 46 i – 10 (-1)

• b) (7i)(5i) = 35  i2 = 35 (-1) = -35
• c) 11i + 13i – 2i = 22i

Conjugate of a Complex Number

It is denoted by z¯

z¯ = a – ib

Modulus of a Complex Number

If z= a+ib 

Then, ।z। = a 2 + b 2

Argand Plane and Polar Plane

A plane just like the XY plane where the complex number a+ib has the coordinates, a and b is called the Argand plane. It is also known as the Gaussian plane. 

The argument of a complex number, z is shown by arg(z)= θ = tan-1(a/b)

arg(z) can also be written as amp(z).                                                   

z is 2nπ + θ is the general value of arg(z) and the length of OP = a 2 + b2

The principal values of the argument lie in the interval (- π, π].

(i) Given x> 0 and y > 0 then, arg (z) = 0

(ii) Given x < 0 and y> 0 then, arg (z) = π -0

(iii) Given x < 0 and y < 0 then, arg (z) = – (π – θ)

(iv) Given x> 0 and y < 0 then, arg (z) = -θz).

If z = a + ib is a complex number, then z in polar form can be written as, 

z = |z| (cos θ + i sin θ) where, θ = arg (z)

If the general value of the argument is 0, then the polar form of z is

z = |z|cos(2nπ+θ)+isin(2nπ+θ), where n is an integer.

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Q.1 If the complex numbers z1, z2 and z3 represent the vertices of an equilateral triangle and |z1| = |z2| = |z3|, then

z1+z2+z3 = 0.

Q.3 If z1=2+i,z2=23i and z3=4+5i, evaluate (i)?Re?(z1?z2z3) (ii)?Im?(z1?z2z3) Marks:4

Ans z1‹…z2 z3=(2+i) (23i ) 4+5i =(2+i) (2+3i) (45i)16+25 =4441+2741iRez1‹…z2 z3 =4441iImz1‹…z2 z3 =2741 Q.4 Find the modulus and the argument of the following: (i)?(2?i)2 (ii)1+i1?i Marks:4

Ans i2i2=414i=34i4i|=9+16=5modulus = 5Since (3,4i) lies in third quadrant, therefore its argument is given by¸

=+tan143ii1+i1i =1+i1i—1+i1+i=11+2i1+1 =i=0+0+i =1=1„modulus =1and argument  is given by=tan110 Q.5 Convert the complex number ?3+i in the polar form and represent it in Argand Plane. Marks:6

Ans Let z=3+i=r(cos¸+isin¸)Comparing real part and imaginary part, we getrcos¸=3 and rsin¸=1On squaring and adding, we getr2cos2¸+sin2¸=32+12r2=4r=2cos¸=32 and sin¸=12The value of ¸ satistying both the equation is by, ¸=56Hence Z=2cos56+isin56

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Faqs (frequently asked questions), 1. what are the applications of complex numbers.

Complex numbers have numerous applications in scientific areas, including fluid dynamics, quantum mechanics, vibration analysis, signal processing, cartography, electromagnetism, control theory, and many more.

2. Define Complex Numbers.

A complex number is defined as a number that can be written as a + ib. A and B are real numbers, and I is an iota.

The value of iota is √-1.

As a result, z (complex number) = a + ib, where a represents the real part and ib represents the imaginary part. Re (z) = a, Im = b (z).

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• NCERT Solutions for Class 11 Maths Chapter 5: Complex Numbers and Quadratic Equations - Exercise 5.3
• NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 5-Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5-Complex Numbers and Quadratic Equations Exercise 5.3 are comprehensive and extensive solutions. They have been prepared by Vedantu’s proficient and skilled subject experts. The Maths NCERT Solutions Class 11 Chapter 5 Exercise 5.3 is accurate and has been prepared as per the CBSE guidelines. On Vedantu get the Exercise 5.3 Class 11 Maths NCERT Solutions in the form of free PDF downloads to prepare better for the examination.

Access NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations.

Exercise 5.3

1. Solve the equation \[{{x}^{2}}+3=0\].

Ans: As, the given equation is a quadratic polynomial. Therefore, we can find the roots of the equation by determining the discriminant.

Hence, we have $a=1$,

Therefore, roots will be –

$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$\Rightarrow x=\frac{\pm \sqrt{0-12}}{2}$

$\Rightarrow x=\frac{\pm 2\sqrt{3}\times \sqrt{-1}}{2}$

$\Rightarrow x=\pm \sqrt{3}i$.

2. Solve the equation \[2{{x}^{2}}+x+1=0\].

Hence, we have $a=2$,

$\Rightarrow x=\frac{-1\pm \sqrt{1-8}}{4}$

$\Rightarrow x=\frac{-1\pm \sqrt{7}\times \sqrt{-1}}{4}$

$\Rightarrow x=\frac{-1\pm \sqrt{7}i}{4}$.

3. Solve the equation \[{{x}^{2}}+3x+9=0\].

$\Rightarrow x=\frac{-3\pm \sqrt{9-36}}{2}$

$\Rightarrow x=\frac{-3\pm 3\sqrt{3}\times \sqrt{-1}}{2}$

$\Rightarrow x=\frac{-3\pm 3\sqrt{3}i}{2}$.

4. Solve the equation \[-{{x}^{2}}+x-2=0\].

Hence, we have $a=-1$,

$\Rightarrow x=\frac{-1\pm \sqrt{1-8}}{-2}$

$\Rightarrow x=\frac{-1\pm \sqrt{7}\times \sqrt{-1}}{-2}$

$\Rightarrow x=\frac{-1\pm \sqrt{7}i}{-2}$.

5. Solve the equation \[{{x}^{2}}+3x+5=0\].

$\Rightarrow x=\frac{-3\pm \sqrt{9-20}}{2}$

$\Rightarrow x=\frac{-3\pm \sqrt{11}\times \sqrt{-1}}{2}$

$\Rightarrow x=\frac{-3\pm \sqrt{11}i}{2}$.

6. Solve the equation \[{{x}^{2}}-x+2=0\].

$b=-1$, and

$\Rightarrow x=\frac{1\pm \sqrt{1-8}}{2}$

$\Rightarrow x=\frac{1\pm \sqrt{7}\times \sqrt{-1}}{2}$

$\Rightarrow x=\frac{1\pm \sqrt{7}i}{2}$.

7. Solve the equation \[\sqrt{2}{{x}^{2}}+x+\sqrt{2}=0\].

Hence, we have $a=\sqrt{2}$,

$c=\sqrt{2}$.

$\Rightarrow x=\frac{-1\pm \sqrt{1-8}}{2\sqrt{2}}$

$\Rightarrow x=\frac{-1\pm \sqrt{7}\times \sqrt{-1}}{2\sqrt{2}}$

$\Rightarrow x=\frac{-1\pm \sqrt{7}i}{2\sqrt{2}}$.

8. Solve the equation \[\sqrt{3}{{x}^{2}}-\sqrt{2}x+3\sqrt{3}=0\].

Hence, we have $a=\sqrt{3}$,

$b=-\sqrt{2}$, and

$c=3\sqrt{3}$.

$\Rightarrow x=\frac{\sqrt{2}\pm \sqrt{2-36}}{2\sqrt{3}}$

$\Rightarrow x=\frac{\sqrt{2}\pm \sqrt{34}\times \sqrt{-1}}{2\sqrt{3}}$

$\Rightarrow x=\frac{\sqrt{2}\pm \sqrt{34}i}{2\sqrt{3}}$.

9. Solve the equation \[{{x}^{2}}+x+\frac{1}{\sqrt{2}}=0\].

$c=\frac{1}{\sqrt{2}}$.

$\Rightarrow x=\frac{-1\pm \sqrt{1-2\sqrt{2}}}{2}$

$\Rightarrow x=\frac{-1\pm \sqrt{2\sqrt{2}-1}\times \sqrt{-1}}{2}$

$\Rightarrow x=\frac{-1\pm \left( \sqrt{2\sqrt{2}-1} \right)i}{2}$.

10. Solve the equation \[{{x}^{2}}+\frac{x}{\sqrt{2}}+1=0\].

$b=\frac{1}{\sqrt{2}}$, and

$\Rightarrow x=\frac{-\frac{1}{\sqrt{2}}\pm \sqrt{\frac{1}{2}-4}}{2}$

NCERT Solutions for Class 11 Maths Chapters

Chapter 1 - Sets

Chapter 2 - Relations and Functions

Chapter 3 - Trigonometric Functions

Chapter 4 - Principle of Mathematical Induction

Chapter 5 - Complex Numbers and Quadratic Equations

Chapter 6 - Linear Inequalities

Chapter 7 - Permutations and Combinations

Chapter 8 - Binomial Theorem

Chapter 9 - Sequences and Series

Chapter 10 - Straight Lines

Chapter 11 - Conic Sections

Chapter 12 - Introduction to Three Dimensional Geometry

Chapter 13 - Limits and Derivatives

Chapter 14 - Mathematical Reasoning

Chapter 15 - Statistics

Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 5 Exercise

Class 11 maths ncert solutions chapter 5 exercise 5.3.

Class 11 is an important year for a student. This is the class when you decide which stream of subjects you want to study. It is the stepping stone towards your future. Hence it is important that you take class 11 very seriously. It may not be your board year, but it helps you prepare the ground for the same. If your concepts are clear and strong in class 11, then learning more complex topics in class 12 will not give you sleepless nights.

NCERT Solutions for Class 11 Maths Chapter 5 deals with various significant topics and subtopics related to Complex Numbers and Quadratic Equations. Some important concepts covered in this chapter include Definition of Complex Numbers, Algebra of Complex Numbers, Argand Plane, Modulus and Argument of Complex Number, Polar Form, De-Moivre’s Theorem, Cube Root of Unity, n nth Root of Unity, Square Root of Complex Numbers, LOCI in Complex Plane, Vectorial Representation of a Complex Number, Important Properties of Complex Numbers, Quadratic Expression, Roots of Quadratic Equations, Nature of Roots, Graph of Quadratic Expression, Solution of Quadratic Inequalities, Maximum and Minimum Value of Quadratic Expression, Theory of Equations, Location of Roots, Maximum and Minimum Values of Rational Expression, Common Roots, Resolution into Two Linear Factors, Formation of A Polynomial Equation and Transformation of Equations.

Exercise 5.3 Maths Class 11 deals with various concepts related to the Algebra of Complex Numbers. Some of the important concepts covered in Class 11 Maths NCERT Solution Exercise 5.3 are Addition of Two Complex Numbers, Difference of Two Complex Numbers, Multiplication of Two Complex Numbers, Division of Two Complex Numbers, Power of I, The Square Roots of a Negative Real Number, and Identities.

CBSE Maths Class 11 NCERT Solutions Chapter 5 Exercise 5.3

NCERT Solutions for Class 1 Maths Chapter 5 Exercise 5.3 Chapter include all possible questions from this section of the chapter. Since the NCERT Exercise follows the CBSE question pattern, solving the Class 11 Exercise 5.3 will give you a fair idea of the kind of question types that are important from an exam point of view.   Class 11 Maths Exercise 5.3 Solutions extensively cover all concepts of the chapter. This ensures that you don’t miss out on anything. 

Vedantu’s Class 11 Maths NCERT Solutions Chapter 5.3 has been prepared in a step by step manner to make understanding of the exercise easier, faster and more effective. They of the finest quality and adhere to all CBSE (Central Board for Secondary Education) and NCERT (National Council of Educational Research and Training) guidelines. Vedantu’s NCERT Solutions have been prepared on the basis of proven and effective study strategies that help you develop a thorough conceptual knowledge of the subject matter quickly and increase your retention rate. These solutions are the shortest, simplest and most logical solutions to the question in the Class 10 Maths Chapter 5 Exercise 5.3. They are just what you need to ace your examinations.

On Vedantu, your entire course content is available in a single place in a well-structured manner on the basis of class, subject, and chapter. This has made learning a very time and effort saving exercise. Vedantu has the perfect resource for solid exam preparation. Vedantu’s expert teachers ensure that Vedantu’s NCERT Solutions focus on complete conceptual clarity. Vedantu gives identical importance to all subjects, all chapters, all topics and all subtopics of a chapter. Vedantu’s Maths NCERT solutions comprise of all solved examples and step by step answers to all exercise questions.

On Vedantu, now you can attend LIVE Master Classes, attend Lectures and LIVE Quizzes, watch Free Conceptual Videos, participate in Ongoing Contests, learn from Vedantu’s LIVE Courses and take the Online Tests. Vedantu is a holistic and effective learning solution that helps you make the best of your audio and visual capabilities to unleash your full potential prepare for your exams with all your might and perform like a star in your exams. Taking Vedantu’s Online Tests give you the confidence to handle the exam question pattern. It helps you become more accustomed to the exam environment and get rid of the fear and anxieties related to exams. These Online Tests let you test the quality of your exam preparation and provoke you to better your preparation. They also boost your self-confidence massively.

An Overview of the Important Topics Covered in Exercise 5.3 of Class 11 Maths NCERT Solutions

Exercise 5.3 of Class 11 Maths NCERT Solutions is mainly based on solving various quadratic equations.

Below are some key learnings from Exercise 5.3 of NCERT Solutions Class 11 Maths. 

The general form of a quadratic equation can be written as ax 2 + bx + c where a, b, and c are the coefficients. 

The quadratic equations (in the real number set) have two real roots and a non-negative discriminant that is b 2 – 4ac ≥ 0. 

The quadratic equations (in the complex number set) have two non-real roots and a negative discriminant value that is b 2 – 4ac < 0. 

This exercise consists of questions based on finding the complex roots of quadratic equations. It also includes multiple examples and problems to explain the use of complex numbers in quadratic equations. These solutions will be helpful for the students to build a deeper understanding of the concepts.

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NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1

NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 Linear Inequalities in English and Hindi Medium updated for CBSE Session 2023-24. The revised solutions of ex. 5.1 class 11th mathematics are given here. It is modified according to rationalised textbooks published by NCERT for academic year 2023-24.

Class 11 Maths Exercise 5.1 Solutions in Hindi and English Medium

• Class 11 Maths Exercise 5.1 in English Medium
• Class 11 Maths Exercise 5.1 in Hindi Medium
• Class 11 Maths NCERT Books in PDF Format
• Class 11 Maths Chapter 5 NCERT Solutions
• NCERT Solutions for Class 11 Maths

Linear inequalities are the advanced version of what you have studied in the previous class linear equations . Earlier you have studied linear equations in one variable and two variables along with some problems where you have been asked to find or prove the solutions.

This gave raised the natural question of what is linear inequality? Well, class 11 Mathematics chapter 5 can describe this as the relationship between two different quantities that are not equal. These inequalities use different mathematical signs like – Less than > and greater than < and more such as greater than and equal to represent the conditions.

Quoting that “Mathematics is the art of saying many things in many different ways” at the beginning of the introductions chapter made it clear that a simple signature stated in inequalities will form the questions. This is visible in 11th Maths NCERT exercise 5.1 that starts with a low difficulty level questions. There you are asked to solve the expression of the single equation. This proceeds to advance case study-based expression where you are given the task to take the expression yourself and then solve it.

The use of signs in inequalities happens when one expression is been compared on the line depending on its value. To make it simple, we can say that when two algebraic expressions are being compared and the result is one expression. This is known as linear inequalities. Passage 5.1 of inequalities are all about recognizing what inequality would look like. The benefit of recognizing the inequality will get you the vision of the correct it or prove it.

Since it is a new topic, which includes symbols of greater than and equal, less than and equal, extends the reach of algebra to a more advanced level. There are various types of such expression in Maths. The inequality one is well explained with a case study of purchasing rice, pen and register. Just read the passage given above the exercise 5.1 of 11th Maths NCERT book.

The passage is easy to understand. After this you will read the actual definition of inequalities. There are different types of inequality expression namely – Double inequalities, Strict inequalities, and slack inequality. This also includes linear inequality in one variable, also linear inequality in two variables.

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NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Ncert solutions class 11 maths chapter 5 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 4.

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at BYJU’S. These NCERT Solutions of Maths help students in solving problems quickly, accurately and efficiently. Also, BYJU’S provides step-by-step solutions for all NCERT problems, thereby ensuring students understand them and clear their board exams with flying colours. The chapter Complex Numbers and Quadratic Equations is categorised under the CBSE Syllabus for 2023-24 and includes different critical Mathematical theorems and formulae. The NCERT textbook has many practice problems to cover all these concepts, which would help students easily understand higher concepts in future. BYJU’S provides solutions for all these problems with proper explanations. These NCERT Solutions from BYJU’S help students who aim to clear their exams even with last-minute preparations. However, NCERT Solutions for Class 11 Maths are focused on mastering the concepts along with gaining broader knowledge.

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Access answers of maths ncert class 11 chapter 5 – complex numbers and quadratic equations.

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Access the exercises of Maths NCERT Class 11 Chapter 5

Exercise 5.1 Solutions 14 Questions Exercise 5.2 Solutions 8 Questions Exercise 5.3 Solutions 10 Questions Miscellaneous Exercise on Chapter 5 Solutions 20 Questions; the summarisation of the topics discussed in Chapter 5 of the Class 11 NCERT curriculum is listed below.

Access NCERT Solutions for Class 11 Maths Chapter 5

Exercise 5.1 Page No: 103

Express each of the complex numbers given in Exercises 1 to 10 in the form a + ib.

1. (5i) (-3/5i)

(5i) (-3/5i) = 5 x (-3/5) x i 2

= -3 x -1 [i 2 = -1]

(5i) (-3/5i) = 3 + i0

2. i 9 + i 19

i 9 + i 19 = (i 2 ) 4 . i + (i 2 ) 9 . i

= (-1) 4 . i + (-1) 9 .i

= 1 x i + -1 x i

i 9 + i 19 = 0 + i0

i -39 = 1/ i 39 = 1/ i 4 x 9 + 3 = 1/ (1 9 x i 3 ) = 1/ i 3 = 1/ (-i) [i 4 = 1, i 3 = -I and i 2 = -1]

Now, multiplying the numerator and denominator by i we get

i -39 = 1 x i / (-i x i)

i -39 = 0 + i

4. 3(7 +  i 7) +  i (7 +  i 7)

3(7 +  i 7) +  i (7 +  i 7) = 21 + i 21 + i 7 + i 2 7

= 21 + i 28 – 7 [ i 2 = -1]

= 14 + i 28

3(7 +  i 7) +  i (7 +  i 7) = 14 + i 28

5. (1 –  i ) – (–1 +  i 6)

(1 –  i ) – (–1 +  i 6) = 1 –  i + 1 –  i 6

(1 –  i ) – (–1 +  i 6) = 2 – i 7

8. (1 –  i ) 4

(1 –  i ) 4 = [(1 –  i ) 2 ] 2

= [1 + i 2 – 2 i ] 2

= [1 – 1 – 2 i ] 2 [ i 2 = -1]

Hence, (1 –  i ) 4 = -4 + 0 i

9. (1/3 + 3 i ) 3

Hence, (1/3 + 3 i ) 3 = -242/27 – 26 i

10. (-2 – 1/3 i ) 3

(-2 – 1/3 i ) 3 = -22/3 – 107/27 i

Find the multiplicative inverse of each of the complex numbers given in Exercises 11 to 13.

Let’s consider z  = 4 – 3 i

= 4 + 3 i  and

|z| 2 = 4 2 + (-3) 2 = 16 + 9 = 25

Thus, the multiplicative inverse of 4 – 3 i  is given by z -1

12. √5 + 3 i

Let’s consider  z  = √5 + 3 i

|z| 2 = (√5) 2 + 3 2 = 5 + 9 = 14

Thus, the multiplicative inverse of √5 + 3 i is given by z -1

Let’s consider  z  = – i

Thus, the multiplicative inverse of – i  is given by z -1

14. Express the following expression in the form of a + ib:

Exercise 5.2 Page No: 108

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

1. z = – 1 – i √3

2. z = -√3 + i

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

Exercise 5.3 Page No: 109

Solve each of the following equations:

1. x 2 + 3 = 0

Given the quadratic equation,

x 2  + 3 = 0

On comparing it with  ax 2  +  bx  +  c  = 0, we have

a  = 1,  b  = 0, and  c  = 3

So, the discriminant of the given equation will be

D =  b 2  – 4 ac  = 0 2  – 4 × 1 × 3 = –12

Hence, the required solutions are

2. 2x 2 + x + 1 = 0

2 x 2  +  x  + 1 = 0

a  = 2,  b  = 1, and  c  = 1

D =  b 2  – 4 ac  = 1 2  – 4 × 2 × 1 = 1 – 8 = –7

3. x 2 + 3x + 9 = 0

x 2  + 3 x  + 9 = 0

a  = 1,  b  = 3, and  c  = 9

D =  b 2  – 4 ac  = 3 2  – 4 × 1 × 9 = 9 – 36 = –27

4. – x 2  +  x  – 2 = 0

– x 2  +  x  – 2 = 0

a  = –1,  b  = 1, and  c  = –2

D =  b 2  – 4 ac  = 1 2  – 4 × (–1) × (–2) = 1 – 8 = –7

5. x 2  + 3 x  + 5 = 0

x 2  + 3 x  + 5 = 0

a  = 1,  b  = 3, and  c  = 5

D =  b 2  – 4 ac  = 3 2  – 4 × 1 × 5 =9 – 20 = –11

6. x 2  –  x  + 2 = 0

x 2  –  x  + 2 = 0

a  = 1,  b  = –1, and  c  = 2

So, the discriminant of the given equation is

D =  b 2  – 4 ac  = (–1) 2  – 4 × 1 × 2 = 1 – 8 = –7

7. √2 x 2  + x  + √2 = 0

√2 x 2  + x  + √2 = 0

a  = √2,  b  = 1, and  c  = √2

D =  b 2  – 4 ac  = (1) 2  – 4 × √2 × √2 = 1 – 8 = –7

8. √3 x 2  – √2 x  + 3√3 = 0

√3 x 2  – √2 x  + 3√3 = 0

a  = √3,  b  = -√2, and  c  = 3√3

D =  b 2  – 4 ac  = (-√2) 2  – 4 × √3 × 3√3 = 2 – 36 = –34

9. x 2  + x  + 1/√2 = 0

x 2  + x  + 1/√2 = 0

It can be rewritten as,

√2 x 2  + √2 x  + 1 = 0

a  = √2,  b  = √2, and  c  = 1

D =  b 2  – 4 ac  = (√2) 2  – 4 × √2 × 1 = 2 – 4√2 = 2(1 – 2√2)

10. x 2  + x /√2 + 1 = 0

x 2  + x /√2 + 1 = 0

D =  b 2  – 4 ac  = (1) 2  – 4 × √2 × √2 = 1 – 8 = -7

Miscellaneous Exercise Page No: 112

2. For any two complex numbers   z 1  and z 2 , prove that

Re (z 1 z 2 )   = Re z 1  Re z 2  – Im z 1  Im z 2

3. Reduce to the standard form.

5. Convert the following into the polar form:

Solve each of the equations in Exercises 6 to 9.

6. 3x 2 – 4x + 20/3 = 0

Given the quadratic equation, 3x 2 – 4x + 20/3 = 0

It can be re-written as: 9x 2 – 12x + 20 = 0

On comparing it with  ax 2  +  bx  +  c  = 0, we get

a  = 9,  b  = –12, and  c  = 20

D =  b 2  – 4 ac  = (–12) 2  – 4 × 9 × 20 = 144 – 720 = –576

7. x 2 – 2x + 3/2 = 0

Given the quadratic equation, x 2 – 2x + 3/2 = 0

It can be re-written as 2x 2 – 4x + 3 = 0

a  = 2,  b  = –4, and  c  = 3

D =  b 2  – 4 ac  = (–4) 2  – 4 × 2 × 3 = 16 – 24 = –8

8. 27x 2 – 10x + 1 = 0

Given the quadratic equation, 27 x 2  – 10 x  + 1 = 0

a  = 27,  b  = –10, and  c  = 1

D =  b 2  – 4 ac  = (–10) 2  – 4 × 27 × 1 = 100 – 108 = –8

9. 21x 2 – 28x + 10 = 0

Given the quadratic equation, 21 x 2  – 28 x  + 10 = 0

On comparing it with  ax 2  +  bx  +  c  = 0, we have

a  = 21,  b  = –28, and  c  = 10

D =  b 2  – 4 ac  = (–28) 2  – 4 × 21 × 10 = 784 – 840 = –56

10. If z 1 = 2 – i , z 2 = 1 + i , find

Given, z 1 = 2 – i , z 2 = 1 + i

12. Let z 1 = 2 – i , z 2 = -2 + i . Find

13. Find the modulus and argument of the complex number.

14. Find the real numbers  x  and  y  if ( x  –  iy ) (3 + 5 i ) is the conjugate of – 6 – 24 i .

Let’s assume z = ( x  –  iy ) (3 + 5 i )

(3x + 5y) – i (5x – 3y) = -6 -24 i

On equating real and imaginary parts, we have

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

Performing (i) x 3 + (ii) x 5, we get

(9x + 15y) + (25x – 15y) = -18 + 120

x = 102/34 = 3

Putting the value of  x  in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

Therefore, the values of  x  and  y are 3 and –3, respectively.

15. Find the modulus of

16. If ( x  +  iy ) 3  =  u  +  iv , then show that

17. If α and β are different complex numbers with |β| = 1, then find

18. Find the number of non-zero integral solutions of the equation |1 – i| x = 2 x.

Therefore, 0 is the only integral solution of the given equation.

Hence, the number of non-zero integral solutions of the given equation is 0.

19. If ( a  +  ib ) ( c  +  id ) ( e  +  if ) ( g  +  ih ) = A +  i B, then show that

( a 2  +  b 2 ) ( c 2  +  d 2 ) ( e 2  +  f 2 ) ( g 2  +  h 2 ) = A 2  + B 2

20. If, then find the least positive integral value of  m .

Thus, the least positive integer is 1.

Therefore, the least positive integral value of  m  is 4 (= 4 × 1).

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations

Chapter 5 of Class 11 Complex Numbers and Quadratic Equations has 3 exercises and a miscellaneous exercise to help the students practise the required number of problems to understand all the concepts. The topics and sub-topics discussed in the PDF of NCERT Solutions for Class 11 of this chapter include 5.1 Introduction We know that some of the quadratic equations have no real solutions. That means the solution of such equations includes complex numbers. Here, we have found the solution of a quadratic equation ax 2 + bx + c = 0 where D = b 2 – 4ac < 0. 5.2 Complex Numbers Definition of complex numbers, examples and explanations about the real and imaginary parts of complex numbers have been discussed in this section. Class 11 Maths NCERT Supplementary Exercise Solutions PDF helps the students to understand the questions in detail. 5.3 Algebra of Complex Numbers 5.3.1 Addition of two complex numbers

5.3.2 Difference of two complex numbers

5.3.3 Multiplication of two complex numbers

5.3.4 Division of two complex number

5.3.5 Power of i

5.3.6 The square roots of a negative real number

5.3.7 Identities

After studying these exercises, students are able to understand the basic BODMAS operations on complex numbers, along with their properties, power of i, square root of a negative real number and identities of complex numbers. 5.4 The Modulus and the Conjugate of a Complex Number The detailed explanation provides the modulus and conjugate of a complex number with solved examples. 5.5 Argand Plane and Polar Representation 5.5.1 Polar representation of a complex number

In this section, it has been explained how to write the ordered pairs for the given complex numbers, the definition of a Complex plane or Argand plane and the polar representation of the ordered pairs in terms of complex numbers.

• A number of the form a + ib, where a and b are real numbers, is called a complex number, “ a” is called the real part, and “ b” is called the imaginary part of the complex number
• z 1 + z 2 = (a + c) + i (b + d)
• z 1 z 2 = (ac – bd) + i (ad + bc)
• For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists a complex number, denoted by 1/z or z–1, called the multiplicative inverse of z
• For any integer k, i 4k = 1, i 4k + 1 = i, i 4k + 2 = – 1, i 4k + 3 = – i
• The polar form of the complex number z = x + iy is r (cosθ + i sinθ)
• A polynomial equation of n degree has n roots.

Disclaimer – 

Dropped Topics – 

5.5.1 Polar Representation of a Complex Number 5.6 Quadratic Equation Example 11 and Exercise 5.3 Examples 13, 15, 16 Ques. 5–8, 9 and 13 (Miscellaneous Exercise) Last three points in the Summary 5.7 Square-root of a Complex Number

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 5

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CBSE Case Study Questions for Class 11 Maths Complex Numbers Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Complex Numbers  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Complex Numbers PDF

Checkout our case study questions for other chapters.

• Chapter 3 Trigonometric Functions Case Study Questions
• Chapter 4 Principle of Mathematical Induction Case Study Questions
• Chapter 6 Linear Inequalities Case Study Questions
• Chapter 7 Permutations and Combinations Case Study Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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Applied Mathematics: Foundations for Class 11 students

Mathematics is already divided into two levels by CBSE: Basic (easy) and Standard (challenging, though not as difficult as Class 10). After Class 10, “Basic” is for students who do not wish to continue studying mathematics, and “Standard” is for those who want to continue

Now, the idea struck that people who don’t want to continue with math or aren’t permitted to do so with Mathematics (code 041), which covers Trigonometry, Vectors, 3D, etc., should still be given the opportunity to comprehend the practical necessity for it, therefore CBSE developed Applied Mathematics (code 241). Any skill topic has a certain job role as its focus. However, the elective course Applied Mathematics focuses on the field of Mathematics with a particular focus on its applications.

Much-needed Applied Mathematics Subject

Students who attend secondary school are better equipped to choose their career opportunities when they graduate. A crucial subject that aids students in selecting their career paths in mathematics. In higher education, mathematics is frequently employed as a supplementary topic in the fields of economics, commerce, social sciences, and many others. According to observations, students who want to study Commerce or Social Science in college may not be able to use the senior secondary mathematics curriculum designed for science topics. With this in mind, the mathematics curriculum for senior secondary classes is expanded to include an additional elective subject with the goal of giving pupils useful math training outside of the physical sciences.

Relevance of Class 11 Applied Mathematics Case Study Questions

• To understand the importance of data in modern society, case studies based on data from the fields of business, economics, psychology, education, biology, and census data will be used.
• Class 11 Applied Mathematics Case Study Questions also strengthen logical thinking abilities such as developing and verifying mathematical arguments, framing instances, and identifying counter examples.
• It motivates Class 11 Applied Mathematics students to do mathematical research and to forge linkages between mathematical concepts and other academic fields.
• Class 11 Applied Mathematics students will gain a solid grasp of descriptive and inferential statistics through the Class 11 Applied Mathematics case study questions, which they may then use to describe, evaluate, and draw conclusions from a given set of data.

Illustrations of Class 11 applied mathematics case study questions

The use of mathematical techniques to solve issues in the actual world is known as applied mathematics. A student’s capacity to apply mathematical ideas to situations that occur in the real world is evaluated using Class 11 applied mathematics case study questions

Below are some illustrations of applied mathematics case study problems for class 11

Class 11 applied mathematics case study question 1

Read the Case study given below and attempt any 4 subparts: A farmer, Ramgarh,  took a bank loan from SBI for repairing his house. But he could not pay the amount on time. This resulted in the accumulation of interest and the amount to pay reached Rs.1,00,000. After a few months, the farmer opened a shop that resulted in enough income and the income increased on a regular basis. So he decided to pay the bank loan in a different manner. The farmer visited the bank. He made an agreement with the bank that he will start paying the amount of Rs.1,00,000 without interest from Jan 2020. In January he will pay Rs.5000 and will increase the payment by Rs. 100 in each month, as shown in the figure.

Answer Key:

Class 11 applied mathematics case study question 2

Read the Case study given below and attempt any 4 sub parts: In a colony, as shown in the following picture, an electric pole has been installed. The pole has been tied by strong wire PQ to support this pole, some electricians were working on the staircase PS.

In the left and right side of the pole, two street lights are fixed at a height of 3m and 4m respectively. These lights are given supply by wires PM and PN. The height OP=5 m and O is the origin. Now answer the following questions:

• x + 15y = -75
• x – 15y + 75 = 0
• x + 5y = 50
• x – 5y = 20
• (b) x – 15y + 75 = 0

Class 11 applied mathematics case study question 3

• (b) 900 ways
• (c) 648 ways
• (d) 100 ways
• (a) 252 ways

Class 11 applied mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

Class 11 applied mathematics case study questions presented here are based on real-life applications of ideas learned in Class 11. These questions will assist students to understand how case study questions will be asked in papers. The questions are intended to assess the student’s comprehension of the principles as well as their ability to apply them to realistic conditions.

What students will study in Class 11 Applied Mathematics?

Students in Class 11 Applied Mathematics will study a number of topics. Topics in Class 11 Applied Mathematics are significant for students who want to career in mathematics or a related discipline. Students will build critical problem-solving abilities in their future studies and professions by studying Class 11 Applied mathematics themes.

CBSE Class 11 Applied Mathematics (Code No. 241)

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